Solving an equation in $\mathbb N$

Question

I am trying to solve the equation $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$ I have made the following progress: 1) $x, y z$ have to be larger than $1$ 2) only one of x, y, z can be $2$; rest should be larger 3) WLOG I have assummed $2\leqslant x\leqslant y\leqslant z$. Knowing this, $x$ has to be smaller than $5$. How to find $y$ and $z$ for all cases of $x \in\{2, 3, 4, 5\}$ to be sure that I am finding all cases? Thanks.

Answer 1

The same sort of argument you have made can be repeated. Clearly if $x=5$ you must have $y=z=5$. If $x=4$ you have $\frac 1y+\frac 1z=\frac 7{20}$ or $20(y+z)=7yz$ and $\frac 1y \ge \frac 7{40}$ so $y \in \{4,5\}$ We get $(4,5,10)$ as the only solution. If $x=3, \frac 1y+\frac 1z=\frac 4{15}$ and $y \in \{3,4,5,6,7\}$. There are not too many cases to try. A spreadsheet and copy down can ease the burden.

Answer 2

Usually this does resort to trial and error: I think there are some other numerical methods but I can't remember them off the top of my head. We can assume $x\le y\le z$.

We have $$\frac1y< \frac3{5}-\frac1x \le \frac2y$$

$$\iff \frac{5x}{3x-5}<y\le \frac{10x}{3x-5}$$

Trial and error:

1. $x=y=2 \Rightarrow z=\frac1{\frac3{5}-\frac12-\frac12}=-\frac5{2}$ so this doesn't work.

...

$n.$ $\quad x=3,y=4...,z=60$ works. etc...

Answer 3

This is not an exhaustive list. But it is at least a partial solution.

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$

We know that $\dfrac{1}{ab}=\dfrac{1}{a(a+b)} + \dfrac{1}{b(a+b)}$

Let $x=2$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{1}{10}.$

We can use $a=2$ and $b=5$ to get

$$\frac{3}{5} = \frac{1}{2}+\frac{1}{14}+\frac{1}{35}.$$

Let $x=3$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{4}{15}.$

We consider $\gcd(a,b)=1$. Then $\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{4}{15}$. It seems $a=1$, $b=3$, and $g=5$ will work. We get

$$\frac{3}{5} = \frac{1}{3}+\frac{1}{5}+\frac{1}{15}.$$

Let $x=4$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{7}{20}.$

We consider $\gcd(a,b)=1$. Then $\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{7}{20}$. No solution there. Next we try $\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{14}{40}$. Then $(a,b,g)=(4,10,1)$ gives us

$$\frac{3}{5} = \frac{1}{4}+\frac{1}{4}+\frac{1}{10}.$$

Let $x=5$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{2}{5}.$

We consider $\gcd(a,b)=1$. Then $\dfrac{1}{ag}+\dfrac{1}{bg}=\dfrac{a+b}{abg}=\dfrac{2}{5}$. Then $(a,b,g)=(1,1,5)$ gives us

$$\frac{3}{5} = \frac{1}{5}+\frac{1}{5}+\frac{1}{5}.$$

Answer 4

Above equation shown below:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}$

Above equation has parametric solution given below:

$(x,y,z) = (4p,12p,5pk)$

Where $p=(5k+3)/(9k)$

For $k=2$, we get:

$(x,y,z)= ((26/9),(78/9),(65/9))$